Let $A_1,A_2$ be the projections of $A$ on the external bisectors of $\angle ACB,\angle ABC,$ resp. The pairs of points $\{ B_1,C_1 \}$ and $\{C_1,C_2 \}$ are defined similarly. Let $D,E,F$ be the midpoints of $BC,CA,AB$ and let $J$ be the incenter of $\triangle DEF.$ $\triangle ACA_1$ is right with circumcenter $E$ $\Longrightarrow$ $\angle AEA_1=2\angle ACA_1=180^{\circ}-\angle ACB$ $\Longrightarrow$ $EA_1 \parallel BC$ $\Longrightarrow$ $A_1 \in EF$ and similarly $A_2 \in EF.$ Now, if $P$ denotes the projection of $J$ on $EF$ (tangency point of incircle (J) of DEF with EF), we get $PA_1=PE+EA_1=PE+EA=\tfrac{1}{2}(p-b)+\tfrac{1}{2}b=\tfrac{1}{2}p$ and similarly $PA_2=\tfrac{1}{2}p.$ Hence by Pythagorean theorem for $\triangle JPA_1,$ we obtain $JA_1=JA_2=\sqrt{PJ^2+{PA_1}^2}=\tfrac{1}{2}\sqrt{r^2+p^2},$ which means that all points $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a same circle with center $J$ and diameter $d=\sqrt{r^2+p^2}.$